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In the single-sinusoid AM voltage model, $m$ is the peak fractional excursion of the signed multiplying factor from its mean value $1$. At $m=0$ there is only the carrier wave; at $m=0.5$ its multiplying factor ranges from $0.5$ to $1.5$; at $m=1$ it just touches zero. For $0\le m\le 1$, the wave envelope extrema areThe last equality follows by adding and subtracting the first two. An oscilloscope estimate uses local carrier peaks near the message extrema and needs enough carrier cycles to approximate those bounds. Consistent local peak-to-peak amplitude readings give the same ratio because their common factor of two cancels. Mixing a peak reading with a peak-to-peak reading does not.

For $A=2\mathrm{V}$ and $m=0.5$, the wave envelope ranges from $1$ to $3\mathrm{V}$, giving $(3-1)/(3+1)=0.5$. These are model values. At $m=1.4$, the signed factor has minimum $-0.4$, but the nonnegative wave envelope reaches zero at its sign changes; the extrema formula would misleadingly report $1$. It cannot measure the true depth of overmodulation from the folded envelope alone.

Coherent demodulation recovers a message by multiplying the received signal by a reference carrier wave of the same frequency and known relative wave phase, then passing the result through a low-pass filter. For the ideal AM model and reference $2\cos ({\omega }_{c}t+\phi )$,A unity-gain ideal low-pass filter passing $0$ and $\Omega$ but rejecting $2{\omega }_c$ and $2{\omega }_c\pm \Omega$ leaves $Aq(t)\cos \phi$. Such a separating cutoff exists because $\Omega <2{\omega }_c-\Omega$. At $\phi =0$, subtraction of $A$ recovers $Am\cos (\Omega t)$ exactly for any finite $m\ge 0$, including overmodulation, provided the transmitter and channel preserve this linear waveform. This is the signed factor, not its absolute value.

A constant phase error scales the recovered message by $\cos \phi$: a $\pi /3$ radian error halves it, and a $\pi /2$ error removes it. A known error with nonzero cosine can be compensated in gain; a frequency mismatch makes the relative wave phase drift and no longer gives constant attenuation. The lecture illustrates synchronous multiplication, filtering and phase error.^[ref]

An envelope detector extracts the slowly varying nonnegative wave envelope of a rapid carrier wave. In a simple electrical version a diode charges a capacitor near positive carrier peaks and an attached resistor discharges it between peaks. When $0\le m<1$, ideal envelope extraction gives $E(t)=A[1+m\cos (\Omega t)]$; removing the constant $A$ leaves $Am\cos (\Omega t)$. This permits recovery without a synchronized carrier reference. Carrier injection thus trades transmitted power for simplicity.^[ref]

A working peak detector needs margins as well as $m<1$: $E_{\min }=A(1-m)$ should comfortably exceed the diode's forward-voltage scale and noise, and the time constant $\tau =RC$ must suppress carrier ripple while following a falling wave envelope. The coarse scale window isA time constant that is too short leaves carrier ripple; one that is too long makes the discharging capacitor fall more slowly than the envelope. In the ideal zero-drop, continuous-tracking approximation, following a falling $E$ requires $E/\tau \ge -E^{\prime }$. For $0<m<1$,Differentiating the ratio locates its maximum at $\cos \theta =-m$. This original slope check explains why operation near $m=1$ is demanding even before overmodulation: the upper bound tends to zero while the need to smooth carrier cycles persists. It is an approximate circuit criterion, not a guarantee in the presence of diode drops, loading or noise. At $m=0$ there is no varying message to track; at $m=1$ ideal envelope extraction remains algebraically possible, but the simple detector has no positive trough margin.

For $m>1$, even an ideal noiseless envelope detector returns $A|q(t)|$. Removing its mean or its original offset cannot restore the discarded sign. A more elaborate receiver can use the retained wave phase through coherent demodulation.